Chapter 13, Problem 102A

Interpretation Introduction

Interpretation: The percentage yield of $\text{FeS}$ is to be determined.

Concept introduction:

A reactant that is completely consumed at the end of a chemical reaction is known as the limiting reagent. Since the reaction cannot proceed without this reagent, the amount of product that can be produced is constrained.

Answer to Problem 102A

  $42\%$

Explanation of Solution

The given reaction is as follows:

  $\text{Fe}\left(s\right)+\text{S}\left(s\right)\to \text{FeS}\left(s\right)$

Amount of $\text{Fe}$ is 25.0 g.

Amount of $\text{S}$ is 32.0 g.

Molar mass of $\text{Fe}$ is $\text{55}\text{.85 }\text{g}/\text{mol}$ .

Molar mass of $\text{S}$ is $32\text{g}/\text{mol}$ .

Now, calculate the molar masses of Fe and S.

  $\begin{array}{c}\text{molar mass of Fe}=\frac{\text{mass in g}}{\text{molar mass}}\\ =\frac{25}{\text{55}\text{.85 }\text{g}/\text{mol}}\\ =\text{0}\text{.447 mol}\end{array}$

  $\begin{array}{c}\text{molar mass of S}=\frac{\text{mass in g}}{\text{molar mass}}\\ =\frac{32.0}{\text{ 32}\text{g}/\text{mol}}\\ =\text{1 mol}\end{array}$

Thus, Fe is a limiting reaction in the above reaction. Thus, $\text{0}\text{.447 mol}$ of Fe reacts to form $\text{0}\text{.447 mol}$ of Fes.

Molar mass of Fes is $87.91\text{g}/\text{mol}$ . Calculate the mass of Fes as follows:

  $\begin{array}{c}\text{mass of FeS}=\text{mol}\times \text{molar mass}\\ =0.447\text{ mol}\times 87.91\text{g}/\text{mol}\\ =\text{39}\text{.29 g}\end{array}$

Thus, the theoretical yield of FeS is $\text{39}\text{.29 g}$ .

But the actual yield of FeS is 16.5 g.

Expression of percentage yield is,

  $\begin{array}{c}\text{percentage yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times 100\\ =\frac{16.5\text{ g}}{39.29\text{ g}}\times 100\\ =41.99\%\\ =42\%\end{array}$

Thus, the percentage yield is $42\%$ .